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Let $f(x)=x^4+81 \in \mathbb{Q}[x]$.

  • Find the splitting field $E$ of $f(x)$ and the extension degree $[E:\mathbb{Q}]$.
  • Find all the fields $L$ with $\mathbb{Q} \leq L \leq E$.

Are the roots of $f(x)$ the following??

$$\pm 3 e^{\frac{\pi i }{4}}, \pm 3 e^{\frac{3 \pi i }{4}}$$

So, is the splitting field $$E=\mathbb{Q}(e^{\frac{\pi i}{4}})$$ ??

Then to find all the fields we have to find first the group $G=Gal(E | \mathbb{Q})$, right??

$Irr(e^{\frac{\pi i}{4}}, \mathbb{Q})=x^4+1$

Roots in $E$: $\omega, \omega^3, \omega^5, \omega^7$, where $\omega=e^{\frac{\pi i}{4}}$

So, we have the following embedding??

$$\sigma_j : \omega \mapsto \omega^{2j-1}, j=1,2,3,4$$

So, is it as followed $$G=\{\sigma_j , j=1,2,3,4 \}$$

The order of the subgroups are the following:

$\#H \mid \#G=4 \Rightarrow \#H=1,2,4$

$\#H=1:H=\{id\}$

$\#H=4:H=G$

$\#H=2: H=\text{ cyclic }$

The order of the elements are:

$ord <\sigma_1>=1$

$ord <\sigma_2>=ord <\sigma_3>=ord <\sigma_4>=2$

Is this correct?? Shouldn't we have found also an element with $ord=4$ ??

So, are the field that we are looking for the following??

$$E^{<\sigma_1>}=E \\ E^{<\sigma_2>}=\mathbb{Q}(\omega+\omega^3) \\ E^{<\sigma_3>}=\mathbb{Q}(\omega^2) \\ E^{<\sigma_4>}=\mathbb{Q}(\omega-\omega^3)$$

Is this correct??

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    $\begingroup$ please don't shout by using double question marks. $\endgroup$ – Jef L Jan 8 '15 at 21:26
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Your reasoning is ok except that you left out $\mathbb Q$ from your final answer. The Galois group is the Klein four-group, which has five (three proper) subgroups and no elements of order $4$ (i.e., it is not cyclic).

Also, you can simplify slightly: $\omega^2=e^{\pi i/2}=i$, $\omega+\omega^3=\sqrt 2\, i$, $\omega-\omega^3=\sqrt 2$. (And so $E$ could also be described as $\mathbb Q[i\sqrt 2]$)

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  • $\begingroup$ Which subgroup have I forgotten?? $\endgroup$ – Mary Star Jan 8 '15 at 21:57
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    $\begingroup$ @HagenvonEitzen, I believe a typo has caused you to omit a comma: the splitting field is $\Bbb Q[i,\sqrt2\,]$. $\endgroup$ – Lubin Jan 8 '15 at 22:05
  • $\begingroup$ How do we get $\mathbb{Q}$ ?? $\endgroup$ – Mary Star Jan 9 '15 at 11:39

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